Let $n \in \mathbb{N}$, $K$ a finite field. Denote by $K[[x]]$ the (profinite) ring of formal power series over $K$. Note that $\text{GL}_n(K[[x]])$ is a profinite group.

Is every closed subgroup of $\text{GL}_n(K[[x]])$ (topologically) finitely generated?

An analogy is that for evey prime number $p$, every closed subgroup of $\text{GL}_n(\mathbb{Z}_p)$ is finitely generated (in fact there is a uniform bound on the number of generators for all of the closed subgroups, and I expect to have such a bound here as well).

  • $\begingroup$ Note that the question is about $K[[x]]$ and not $K((x))$. $\endgroup$
    – Pablo
    Commented Aug 9, 2014 at 17:35
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    $\begingroup$ This is false for $n=1$ and the entire group. It suffices to show that for the local field $F = K(\!(x)\!)$, $F^{\times}$ has infinitely many non-trivial continuous homomorphisms to $\mathbf{Z}/p\mathbf{Z}$. By local class field theory, it is the same to say that ${\rm{H}}^1(F, \mathbf{Z}/p\mathbf{Z})$ is infinite. By the Artin-Schreier sequence, this cohomology is the same as the continuous $\mathbf{F}_p$-dual of the pro-$p$ group $F/\wp(F)$ where $\wp(f)=f^p-f$. By considering polar parts, $F/\wp(F)$ is infinite. $\endgroup$
    – user27920
    Commented Aug 9, 2014 at 17:36
  • $\begingroup$ Since $K(\!(x)\!)^{\times} = x^{\mathbf{Z}} \times K[\![x]\!]^{\times}$ as topological groups, to show the answer is negative for $n=1$ and the whole group it suffices to show that $K(\!(x)\!)^{\times}$ has infinitely many continuous homomorphisms to $\mathbf{Z}/p\mathbf{Z}$, as is done in my preceding comment. And then via determinant it fails (on ${\rm{GL}}_n(K[\![x]\!])$) for all $n$. The question should perhaps be posed for SL$_n$. $\endgroup$
    – user27920
    Commented Aug 9, 2014 at 17:44
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    $\begingroup$ I would gladly accept this as an answer. Could you please elaborate more on the connection between $K((x))$ and $K[[x]]$ and how the solution for the former gives a solution for the latter? $\endgroup$
    – Pablo
    Commented Aug 9, 2014 at 18:21
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    $\begingroup$ An explicit indication of infinitely many polar parts that are not equal in $F/\wp(F)$ is mentioned in math.stackexchange.com/questions/353928/…. $\endgroup$
    – KConrad
    Commented Aug 10, 2014 at 6:39

1 Answer 1


This question has a negative answer is many respects. Firstly, there are simple constructions in the commutative case. Namely, the additive group $K[\![x]\!]$ is an infinite dimensional $\mathbb F_p$-vector space and thus not finitely generated. Less simply, the multiplicative group $K[\![x]\!]^\times$ has infinitely many homomorphisms to $\mathbb F_p$. That is, the quotient by the group of $p$-th powers is infinite dimensional. The group of $p$-th powers is easy to calculate because the $p$-th power is Frobenius, a ring homomorphism: it is the power series of the form $\sum a_n t^{np}$, $a_0\ne0$. Thus the elements $1+t^n$, for $n$ not divisible by $p$, forms a linearly independent sequence in the $\mathbb F_p$-vector space of the mod $p$ quotient. Class field theory relates the slightly bigger group of the mod $p$ quotient of $K((x))^\times$ with the abelianized Galois group of $K((x))$ and with the quotient of the additive group by Artin-Schreier map $f\mapsto f^p-f$.

Thus for any $n$ the whole group $\rm{GL}_n(K[\![x]\!])$ is not finitely generated because it surjects via the determinant to the multiplicative group, which is not. So one might restrict to $\rm{SL}_n$. But this contains subgroups that are not semisimple, such as $\mathbb G_a$, $\mathbb G_m$, $\rm{GL}_{n-1}$, and the Borel subgroup. None of these are finitely generated, but all for the same commutative reason. But this is really cheating: clearly the "correct" question should get away from the silly constructions that one can make with ease in the commutative setting, so one ought to replace GL$_n$ with semisimple groups $G$ over $O = K[\![x]\!]$.

Likewise, if $G$ is not simply connected then more commutative silliness gets in the way. For example, the determinant induces ${\rm{PGL}}_n(O) \twoheadrightarrow O^{\times}/(O^{\times})^n$, so for $n$ divisible by $p$ we have the same problem once again (as $O^{\times}/(O^{\times})^p$ is a quotient, and it has infinitely many continuous homomorphisms to $\mathbf{Z}/p\mathbf{Z}$ as we have seen above).

So finally it seems that the version of the question not easily falsified by commutative tricks is for $G$ to be a semisimple $O$-group that is simply connected and to consider the open subgroups of $G(O)$ (which are of course automatically closed). It is a general fact (via deformation theory of semisimple groups) that any such $G$ is the scalar extension along $K \rightarrow O$ of its reduction, though you might wish to just consider only such $G$ without knowing it is the most general case. By coming from $K$ we see that $G$ arises from a "constant" $K$-group over the global ring $K[x]$. One can then use strong approximation for simply connected groups over global fields (such as $K(x)$) and results of Behr on finite generation of $S$-arithmetic groups when the sum of local ranks at $v \in S$ is at least 3) to deduce that $G(O)$ contains a dense finitely generated subgroup, namely an $S$-arithmetic group for suitable $S$, so $G(O)$ is topologically finitely generated. Moreover, any open subgroup of $G(O)$ is defined by a "congruence condition", so one can still find $S$-arithmetic groups dense in such open subgroups, and hence they're again topologically finitely generated.

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    $\begingroup$ Relevant to the last paragraph: an open subgroup of a finitely generated profinite group is finitely generated. This is closely related to: a finite index subgroup of a finitely generated group is finitely generated. $\endgroup$ Commented Aug 10, 2014 at 17:01

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